1. 1 STAT 500 – Statistics for Managers STAT 500 Statistics for Managers

2. 2 STAT 500 – Statistics for Managers Agenda for this Session Normal Distribution

3. 3 STAT 500 – Statistics for Managers Probability Density Function Note that f(x) is not probability, i.e., f(x)  P(X = x)

4. 4 STAT 500 – Statistics for Managers Continuous Random Variable: Values from Interval of Numbers Absence of Gaps Continuous Probability Distribution: Distribution of a Continuous Variable Most Important Continuous Probability Distribution: the Normal Distribution

5. 5 STAT 500 – Statistics for Managers The Normal Distribution ‘Bell Shaped’ ‘Bell Shaped’ Symmetrical Symmetrical Mean, Median and Mean, Median and Mode are Equal Random Variable has Random Variable has Infinite Range Mean Median Mode X f(X) 

6. 6 STAT 500 – Statistics for Managers Many Normal Distributions Varying the Parameters  and , we obtain Different Normal Distributions. There are an Infinite Number

7. 7 STAT 500 – Statistics for Managers Infinitely Many Normal Distributions Means Infinitely Many Tables to Look Up!

8. Z Z Z = 0.12 Z.00.01 0.0.0000.0040.0080.0398.0438 0.2.0793.0832.0871 0.3.0179.0217.0255.0478.02 0.1. 0478 Standardized Normal Probability Table (Portion)  = 0 and  = 1 Probabilities Shaded Area Exaggerated Standardized Normal Distribution

9. Z  = 0  Z = 1.12 Standardizing Example Normal Distribution Standardized Normal Distribution X  = 5  = 10 6.2 Shaded Area Exaggerated

10. 0  = 1 -.21 Z.21 Example: P(2.9 < X < 7.1) =.1664 Normal Distribution.1664.0832 Standardized Normal Distribution Shaded Area Exaggerated 5  = 10 2.97.1X

11. Z  = 0  = 1.30 Example: P(X  8) =.3821 Normal Distribution Standardized Normal Distribution.1179.5000.3821 Shaded Area Exaggerated. X  = 5  = 10 8

12. Z.000.2 0.0.0000.0040.0080 0.1.0398.0438.0478 0.2.0793.0832.0871.1179.1255 Z  = 0  = 1.31 Finding Z Values for Known Probabilities.1217.01 0.3 Standardized Normal Probability Table (Portion) What Is Z Given P(Z) = 0.1217? Shaded Area Exaggerated.1217

13. Z  = 0  = 1.31 X  = 5  = 10 ? Finding X Values for Known Probabilities Normal DistributionStandardized Normal Distribution.1217 Shaded Area Exaggerated X 8.1  Z  = 5 + (0.31)(10) =

14. Applications You are the stores manager of Giant Super Store. The daily demand for delicate chocolate bars is normally distributed with a mean of 1000 and a standard deviation of 100. You stock the item every day and the unsold bars are trashed at the end of the day. (i) What is the probability that you face a stock out when you stock 1200 bars? (ii) What is the probability that you end up trashing 300 or more bars in a given day when you stock 1200 bars? (iii) How much do you stock to avoid a stock out of 1% or less?

15. (i) Mean = 1,000; Standard Deviation = 100; Distribution: Normal P (Stock out) = P (Demand >= 1200) = P (Demand >= (1200-1000) / 100) = 0.5 – 0.4772 = 0.0228 or 2.28% is the probability of stock out when the stock is 1200. Stock out probability

16. Remember, the store trashes the unsold chocolate bars. They have a stock 0f 1,200. They trash 300 or more when the demand id 900 or less. It is indirectly asking the probability that the demand is less than or equal to 900. Probability of trashing We know that the demand is more than 1000 is 0.5. P(demand is between 900 and 1000 is) = probability (Demand >= (900 – 1000) / 100 ) on standard normal = 0.3413 (From tables) Probability that the demand is 900 or less : 0.5 – 0.3413 = 0.1587 Probability of trashing 300 or more is 0.1587

17. (iii) Stock out probability is 0.01 (or less) Locate the value corresponding to an area of 0.49. The value for standard deviation is 2.33 corresponding to a standard normal distribution. Therefore, we have to stock 2.33 times 100 (the standard deviation) over and above the mean to avoid stock-out (beyond 1%). Probability of Stock-out, 0.01 Please think of it this way. You can stock 1000 end up with a stock out of 50%. To avoid stock- out, you need to stock more. How much more depends on the risk (stock-out probability) and the standard deviation of the distribution. In this case, the risk is 1% and the standard deviation is 100. The risk corresponds to a standard deviation of 2.33.

18. 18 STAT 500 – Statistics for Managers Agenda for this Session Normal Distribution